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mariarad [96]
3 years ago
7

What happens when any ray of light strikes a concave mirror, after passing through its focal point?

Physics
2 answers:
nekit [7.7K]3 years ago
6 0

Answer:

Option (D)

Explanation:

The rules to draw the ray diagrams for concave mirror are given below:

1. When a ray of light incident on the pole of concave mirror at an angle of incidence, then after reflection, the angle of incidence is equal to the angle of reflection.

2. When incident ray of light passes through the centre of curvature of the mirror, then the reflected ray retraces its path.

3. When the incident ray of light passing through the focus of the concave mirror, then after reflection the reflected ray becomes parallel to the principal axis and vice versa.

baherus [9]3 years ago
5 0
D) The reflected ray travels parallel to the principal axis
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What is tan -1 (0.52)?
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A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

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3 years ago
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anzhelika [568]

Answer:

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